NP-Complete
Question 1 |
NP-Complete. | |
solvable in polynomial time by reduction to directed graph reachability. | |
solvable in constant time since any input instance is satisfiable. | |
NP-hard, but not NP-complete. |
Question 1 Explanation:
Note: Out of Syllabus.
Question 2 |
Which of the following statements are TRUE?
1. The problem of determining whether there exists a cycle in an undirected graph is in P. 2. The problem of determining whether there exists a cycle in an undirected graph is in NP. 3. If a problem A is NP-Complete, there exists a non-deterministic polynomial time algorithm to solve A.
1, 2 and 3 | |
1 and 2 only | |
2 and 3 only | |
1 and 3 only |
Question 2 Explanation:
Note: Out of syllabus.
1. Detecting cycle in a graph using DFS takes O(V+E)=O(V2)
Here, for complete graph E<= V2. So, It runs in polynomial time.
2. Every P-problem is NP because P subset of NP (P ⊂ NP)
3. NP – complete ∈ NP.
Hence, NP-complete can be solved in non-deterministic polynomial time.
1. Detecting cycle in a graph using DFS takes O(V+E)=O(V2)
Here, for complete graph E<= V2. So, It runs in polynomial time.
2. Every P-problem is NP because P subset of NP (P ⊂ NP)
3. NP – complete ∈ NP.
Hence, NP-complete can be solved in non-deterministic polynomial time.
Question 3 |
Which of the following problems is not NP-hard?
Hamiltonian circuit problem | |
The 0/1 Knapsack problem | |
Finding bi-connected components of a graph | |
The graph colouring problem |
Question 3 Explanation:
Note: Out of syllabus.
There are 3 questions to complete.